Q2 Flashcards
1
Q
Needs/purpose for electrical isolation
A
- Isolate low dc voltages from high grid voltages (improving safety for user)
- Transformer allow us to convert voltage levels between primary and secondary to reduce I and V stress of power devices (when large dc-dc conversion needed -i.e. using flyback instead of boost)
- To break loops and separate parts of circuit sensitive to noise.
- To have different reference potentials with respect to the input, output V is fixed but output can be shifted or inverted.
- transformer of isolated converters operate at high frequencies, which reduces its size and weight->(high power density)
- More efficient than linear power supplies, dc-dc isolated converters operate the devices in the ohmic region not in active region.
2
Q
Problems caused by leakage inductance
A
- Energy stored by leakage L must be absorbed by power devices, increasing stress on the devices and power losses
- Additional components needed to dissipate leakage energy and protect power devices
3
Q
Disadvantage of HF transformers
A
- Not perfect coupling between windings can be achieved, resulting in a leakage L.
4
Q
Alternatives to reduce leakage L
A
- Good transformer design
- Passive snubbers (RCD)
- Active snubbers
- Using leakage L for the operation of the converter.
5
Q
Explain general ideas and limitations of good design of transformer
A
- Maximize coupling between primary and secondary
- Minimizing distance between primary and secondary windings
- i.e. in flyback transformer we can use this structure where primary winding is split into 2 layers to increase coupling
- Leakage L cannot be removed completely
6
Q
Explain general ideas of passive snubers (RCD in flyback)
A
- Snubbers are used to absorb/dissipate leakage energy in a resistive element
- RCD snubber in flybacks limits voltage stress in the MOSFET and dissipates leakage energy in the R
7
Q
explain general idea of active snubber in flyback
A
- Active snubber allow recycling the leakage energy and reduce power losses
- i.e. in flyback when MOSFET Q1 is turned off, leakage I flows through body diode of Q2 storing energy in clamp C, then clamp C starts to resonate with leakage L, and MOSFET Q2 is turned on so that leakage I changes direction and recycles energy to the output.
8
Q
Explain idea of using leakage L in resonant converters
A
- Leakage L can be used for the operation of the converter, known as resonant converter
- By using resonant converter it’s possible to achieve zero-voltage or zero-current switching reducing switching losses and losses related to leakage L.
9
Q
Q3. flyback CCM main ideas
A
- In CCM characteristics
- Minimum primary L
- Voltage across primary side inductor ON/OFF state, determines rate of change of magnetizing I
- Voltage across MOSFET/diode ON/OFF state
- Current of MOSFET/diode ON/OFF state
10
Q
Characteristics and condition for CCM operation of flyback
A
- Only a part of energy stored in flyback transformer delivered to output, some energy remains when next cycle starts
- To operate in CCM, Lp should be large enough to ensure magnetizing I never reach zero in the OFF interval
11
Q
Definition of DCM operation of flyback
A
- All energy stored in transformer is transferred to the load in the off period
- Magnetizing I goes to zero
12
Q
Advantages of DCM
A
- Lp value is smaller reducing size of transformer
- Soft switching of secondary diode, I decreases with slope -Vo/Ls and not abruptly, it might lead to less recovered charge and losses.
- Turn-on of MOSFET is softer since I increases from zero with slope Vi/Lp and not abruptly, it might lead to less switching losses.
- Possibility of implement QR resonant flyback due to resonance between Lp and output C of MOSFET
13
Q
Disadvantages of DCM
A
- Larger ripple and peak currents
- Higher MOSFET turn off losses due to higher peak I
- Higher leakage energy that has to be dissipated due to higher peak I
14
Q
Advantages of CCM
A
- Smaller ripple I and peak I
- Lower MOSFET turn-off losses
15
Q
Disadvantages of CCM
A
- Inductance value for CCM is higher, leading to bigger transformer
- Hard-switching of diode, leading to increased reverse-recovery losses due to high di/dt and reverse peak I.
- Needs slope compensation for duty cycles higher than 50%
16
Q
Steps to find voltage gain calculation in CCM
A
- Find voltage across Lp when switch is ON
- Find voltage across Lp when swtich is OFF
- Apply second-volt balance of Lp, adding areas under curve of Vl and equating them to zero.
17
Q
Steps to find voltage gain in DCM and observations
A
- Consider all energy stoed in Lp is transferred to output during each switching period
- Calculate the average input power dividing energy stored in Lp by Ts
- Find primary peak I multiplying slope Vin/Lp by the time the switch is ON (DTs)
- Voltage gain expression is more complex and depends on fs, output load, Lp.
18
Q
Advantages of flyback for multiple outputs and limitations
A
- Flyback can have multiple outputs from a single input source.
- Output voltages will be proportional to the turns ratio of each winding.
- Ideally we only need to regulate 1 output and the other will scale by the number of turns.
- Parasitic elements affect load regulation of unregulated outputs.
19
Q
Explain how cross regulation is affected in a multiple output flyback
A
- Occurs when 1 or more outputs are at minimum or zero load, and 1 output is at maximum load.
- When MOSFET is turned off, leakage L in the primary induces a voltage spike common to all windings.
- With no output load the C is charged to the peak voltage, until the diode stops conducting and C cannot discharge.
- If load of 1 output keeps increasing, voltage spike caused by leakage L increases and voltage of output with no load increase.
20
Q
Control techniques for multiple output flyback and general explanation
A
- AC stacked windings and combined feedback
- Output 1 voltage will be proportional to (m+n) and output 2 voltage will be proportional to m
- combine the outputs connecting them with a R to reference pin of shunt regulator
- Regulator controls combination of outputs with only one feedback point, so no output will be perfectly regulated
- i.e. if W1=0.9 and W2=0.1, output 1 will be 9 times more important than output 2, and will have better regulation.
21
Q
Explain procedure to calculate R of combined feedback method
A
- We assign an importance/weight to each output W1 and W2
- total I through R0 will be contribution of each output multiplied by the weight
- we know that i1 will be difference between output and Vref times R1, and solve for R1
- We do the same for R2
